Subject: Re: [xsl] position within same nodes From: David Carlisle <davidc@xxxxxxxxx> Date: Thu, 19 Jul 2001 21:20:32 +0100 |
What is the XPath to get the position of a given node relative to its siblings of the same name? Eg, <blah> <argh/> <meep/> <meep/> <stuff/> <meep/> </blah> I'm looking for an XPath statement that would return 1 for the first meep, 2 for the second, and 3 for the 3rd meep, rather than 2, 3, and 5 (which is what plain position() returns). I'm trying to count the number of siblings and subtract the number of siblings of the same type, then subtract that from position(), but I can't get either of the first two parts there to work either. Does anyone have any suggestions? you are under a misaprehention about position() check the archives of this list. As stated many times nodes do not have a position() intrinsic to themselves, they just have a position in the current node list. so the same node will have different positions at different times, If you were to go select="meep" you would have a node lists of just th emeep elements and so they would have position() of 1 2 3. If you go select="*" you pick up all the elements and so the meeps have positions of 2 3 and 5. If you use the default value of apply-templates and you haven't stripped white space then the current node list would also have text nodes and the meeps would have position of 4 6 and 10 (if I counted right) If you want to know th eposition in the tree, count(previous-sibling::meep) might be what you are after, it all depends. David _____________________________________________________________________ This message has been checked for all known viruses by Star Internet delivered through the MessageLabs Virus Scanning Service. For further information visit http://www.star.net.uk/stats.asp or alternatively call Star Internet for details on the Virus Scanning Service. XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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